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How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R

(I came across this while solving a problem)

w = (1/LC - R

^{2}/4L^{2})^{1/2}(I came across this while solving a problem)

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- Thread starter zorro
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w = (1/LC - R

(I came across this while solving a problem)

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- #2

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You solve the differential equation for kirchoff's loop rule.

- #3

tiny-tim

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How does this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R^{2}/4L^{2})^{1/2}

(I came across this while solving a problem)

Well, without knowing anything about circuits, we can guess that it probably has something to do with the fact that that obviously looks like a

of Lx^{2} + Rx + 1/C = 0 …

can you get a differential equation that looks like that, maybe for current?

- #4

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w = (-R +/- (R

I did not get any differential equation.

- #5

tiny-tim

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I got the same equation for w in place of x. On solving,

w = (-R +/- (R^{2}-4L/C)^{1/2})/2L

Are you sure? that's not the same ω as in your original post …

How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R^{2}/4L^{2})^{1/2}

I suspect that the R/2L is a separate exponential part

(my differential equation in I was from combining CV

- #6

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(my differential equation in I was from combining CV_{C}= Q (so CdV_{C}/dt = I), V_{R}= IR, V_{L}= LdI/dt, and d/dt(V_{C}+ V_{R}+ V_{L}) = 0)

That will give a second order differential equation, right?

Your expression does not contain any w.

- #7

tiny-tim

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part of that solution will be harmonic, and ω will be the frequency of the harmonic part.

- #8

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I got

I/C + RdI/dt + Ld^{2}I/dt^{2} = 0

How to solve this?

I/C + RdI/dt + Ld

How to solve this?

- #9

tiny-tim

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(just got up :zzz: …)

I got

I/C + RdI/dt + Ld^{2}I/dt^{2}= 0

How to solve this?

Don't you know the standard method for these equations? …

write D = d/dt, so the equation becomes

(LD

which you can factor to

(D - R/2L + i√(1/LC + R

carry on from there

- #10

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Its d^{2}I/dt^{2} not (dI/dt)^{2}

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tiny-tim

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D

- #12

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I have solved questions in D.E. like -

1 + (dy/dx)

by taking dy/dx = p

Do we use p

- #13

tiny-tim

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no …

that p equation has an xdy/dx,

the D is for equations with just dy/dx

that p equation has an xdy/dx,

the D is for equations with just dy/dx

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